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a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)}{\sqrt{a}-1}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: P<1
=>P-1<0
=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
=>\(\dfrac{a+2}{\sqrt{a}-1}< 0\)
=>căn a-1<0
=>0<=a<1
c: Khi a=19-8căn 3=(4-căn 3)^2 thì \(P=\dfrac{19-8\sqrt{3}+4-\sqrt{3}+1}{4-\sqrt{3}-1}=\dfrac{24-9\sqrt{3}}{3-\sqrt{3}}=\dfrac{15-\sqrt{3}}{2}\)
a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
hay 0<a<1
\(a,P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\left(-1< a< 1\right)\\ P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{2+\sqrt{\left(1-a\right)\left(1+a\right)}}\\ P=\sqrt{1-a}\\ b,a=\dfrac{24}{49}\Leftrightarrow1-a=\dfrac{25}{49}\\ \Leftrightarrow P=\sqrt{1-a}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\\ c,P=2\Leftrightarrow1-a=4\Leftrightarrow a=-3\left(ktm\right)\Leftrightarrow a\in\varnothing\)
`P=(sqrta+3)/(sqrta-2)-(sqrta-1)/(sqrta+2)+(4sqrta-4)/(4-a)`
`đk:x>=0,x ne 4`
`P=(a+5sqrta+6-a+3sqrta-2-4sqrta+4)/(a-4)`
`=(4sqrta+8)/(a-4)`
`=4/(sqrta-2)`
`b)a=9`
`=>P=4/(3-2)=4`
a) Ta có: \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
\(=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\dfrac{4}{\sqrt{a}-2}\)
b) Thay a=9 vào P, ta được:
\(P=\dfrac{4}{\sqrt{9}-2}=\dfrac{4}{3-2}=\dfrac{4}{1}=4\)
Vậy: khi a=9 thì P=4
Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1-2\sqrt{a}+a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b) Để \(P< \dfrac{1}{2}\) thì \(P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-1\right)}{2\sqrt{a}}-\dfrac{\sqrt{a}}{2\sqrt{a}}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{a}-2-\sqrt{a}}{2\sqrt{a}}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{2\sqrt{a}}< 0\)
mà \(2\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2< 0\)
\(\Leftrightarrow\sqrt{a}< 2\)
hay a<4
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)
Vậy: Để \(P< \dfrac{1}{2}\) thì \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)
a) ĐK : a > = 0 ; a + 1 # 0 ; căn a - 1 # 0 <=> a >= 0 ; a # 1
Mẫu số = 1/(căn a - 1) - 2.căn a / [(a.căn a - a) + (căn a - 1)]
= 1/(căn a - 1) - 2.căn a / [(a + 1)(căn a - 1)]
= [(a + 1) - 2.căn a] / [(a + 1)(căn a - 1)]
= (căn a - 1)^2 / [(a + 1)(căn a - 1)]
= (căn a - 1) / (a + 1)
Tử số = (a + 1 + căn a) / (a + 1)
=> B = [(a + 1 + căn a) / (a + 1)] : [(căn a - 1) / (a + 1)]
= (a + căn a + 1) / (căn a - 1)
b) a = 19 - 8.căn3 = 16 - 2.4.căn3 + 3 = (4 - căn3)^2
B = ( 19 - 8.căn3 + 4 - căn3 + 1) / (4 - căn3 - 1)
= (24 - 9.căn3) / (3 - căn3)
= [(24 - 9.căn3)(3 + căn3)] / [(3 - căn3)(3 + căn3)]
= (45 - 3.căn3) / 6 = (15 - căn3) / 2
NHỚ TK MK NHA , MK ĐANG ÂM ĐIỂM